Inspired by Satyajit Mohanty

Calculus Level 5

1 − 1 2 + 1 3 − 1 5 + 1 6 − 1 7 + 1 9 − 1 10 + 1 11 − 1 13 + 1 14 − 1 15 + 1 17 − … 1-\frac { 1 }{ 2 } +\frac { 1 }{ 3 } -\frac { 1 }{ 5 } +\frac { 1 }{ 6 } -\frac { 1 }{ 7 } +\frac { 1 }{ 9 } -\frac { 1 }{ 10 } +\frac { 1 }{ 11 } -\frac { 1 }{ 13 } +\frac { 1 }{ 14 } -\frac { 1 }{ 15 } +\frac { 1 }{ 17 } -\dots If the above expression can be expressed as ( A B − C ) π D E \dfrac { (A\sqrt { B } -C){ \pi }^{ D } }{ E } for positive integers A , B , C , D A,B,C,D and E E . Find the minimum value of A × B × C × D × E A\times B \times C \times D \times E

Inspired by Problem 1 , Problem 2 , Problem 3 .


The answer is 32.

Rajorshi Chaudhuri by Rajorshi Chaudhuri , 23, India

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1 solution

Let the given expression be S S , then,

S = 1 − 1 2 + 1 3 − 1 5 + 1 6 − 1 7 + 1 9 − 1 10 + 1 11 − 1 13 + 1 14 − 1 15 + 1 17 − . . . = 2 ( 1 2 − [ 1 2 + 1 − 1 ] 1 2 + 1 3 2 + 1 4 ( 0 ) − 1 5 2 + [ 1 2 + 1 − 1 ] 1 6 − 1 7 2 + 1 8 ( 0 ) + 1 9 2 − . . . ) = 2 ( 1 2 + 1 2 + 1 3 2 + 1 4 ( 0 ) − 1 5 2 − 1 6 − 1 7 2 + 1 8 ( 0 ) + 1 9 2 − . . . ) − ( 1 + 2 ) ( 1 2 − 1 6 + 1 10 − . . . ) = 2 ( sin ⁡ π 4 + sin ⁡ π 2 2 + sin ⁡ 3 π 4 3 + sin ⁡ π 4 + sin ⁡ 5 π 4 5 + sin ⁡ 3 π 2 6 + sin ⁡ 7 π 4 7 + . . . ) − 1 + 2 2 ( 1 − 1 3 + 1 5 − . . . ) = 2 ℑ ( ∑ n = 1 ∞ e i n π 4 n ) − 1 + 2 2 ∑ n = 0 ∞ ( − 1 ) n 2 n + 1 = 2 ℑ ( − ln ⁡ ( 1 − e i π 4 ) ) − 1 + 2 2 ( π 2 − cot ⁡ − 1 ( 1 ) ) = − 2 ℑ ln ⁡ ( 1 − cos ⁡ π 4 − i sin ⁡ π 4 ) − 1 + 2 2 ( π 2 − π 4 ) = − 2 ℑ ln ⁡ ( 1 − 1 2 − i 1 2 ) − 1 + 2 2 ( π 4 ) = − 2 ℑ ln ⁡ ( 2 − 2 e i tan ⁡ − 1 − 1 2 − 1 ) − ( 1 + 2 ) π 8 = − 2 ℑ ln ⁡ ( 2 − 2 e − i 3 π 8 ) − ( 1 + 2 ) π 8 = 3 2 π 8 − ( 1 + 2 ) π 8 = ( 2 2 − 1 ) π 8 \begin{aligned} S & = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{5} + \frac{1}{6} - \frac{1}{7} + \frac{1}{9} - \frac{1}{10} + \frac{1}{11} - \frac{1}{13} + \frac{1}{14} - \frac{1}{15} + \frac{1}{17} -... \\ & = \small \sqrt{2}\left(\frac{1}{\sqrt{2}} - \color{#3D99F6}{\left[\frac{1}{\sqrt{2}} +1 - 1 \right] \frac{1}{2}} + \frac{1}{3\sqrt{2}}+ \frac{1}{4}(0) - \frac{1}{5\sqrt{2}} + \color{#3D99F6}{\left[\frac{1}{\sqrt{2}} +1 - 1 \right]\frac{1}{6}} - \frac{1}{7\sqrt{2}} + \frac{1}{8}(0) + \frac{1}{9\sqrt{2}}- ...\right) \\ & = \small \sqrt{2}\left(\frac{1}{\sqrt{2}} + \color{#3D99F6}{\frac{1}{2}} + \frac{1}{3\sqrt{2}}+ \frac{1}{4}(0) - \frac{1}{5\sqrt{2}} - \color{#3D99F6}{\frac{1}{6}} - \frac{1}{7\sqrt{2}} + \frac{1}{8}(0) + \frac{1}{9\sqrt{2}}- ...\right) - \color{#3D99F6}{\left(1+\sqrt{2}\right)\left( \frac{1}{2} - \frac{1}{6} + \frac{1}{10} - ... \right)} \\ & = \small \sqrt{2}\left(\sin \frac{\pi}{4} + \frac{\sin \frac{\pi}{2}}{2} + \frac{\sin \frac{3\pi}{4}}{3}+ \frac{\sin \pi}{4} + \frac{\sin \frac{5\pi}{4}}{5} + \frac{\sin \frac{3\pi}{2}}{6} + \frac{\sin \frac{7\pi}{4}}{7} + ...\right) - \frac{1+\sqrt{2}}{2}\left( 1 - \frac{1}{3} + \frac{1}{5} - ... \right) \\ & = \sqrt{2}\Im \left( \color{#3D99F6}{\sum_{n=1}^\infty \frac{e^{i\frac{n\pi}{4}}}{n}} \right) - \frac{1+\sqrt{2}}{2}\color{#D61F06}{\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}} \\ & = \sqrt{2}\Im \left( \color{#3D99F6}{-\ln \left(1-e^{i\frac{\pi}{4}}\right)} \right) - \frac{1+\sqrt{2}}{2}\left(\color{#D61F06}{\frac{\pi}{2}-\cot^{-1} (1)}\right) \\ & = \color{#3D99F6}{-}\sqrt{2}\Im \color{#3D99F6}{\ln \left(1-\cos \frac{\pi}{4}-i\sin \frac{\pi}{4} \right)} - \frac{1+\sqrt{2}}{2}\left(\color{#D61F06}{\frac{\pi}{2}-\frac{\pi}{4}}\right) \\ & = \color{#3D99F6}{-}\sqrt{2}\Im \color{#3D99F6}{\ln \left(1-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}} \right)} - \frac{1+\sqrt{2}}{2}\left(\color{#D61F06}{\frac{\pi}{4}}\right) \\ & = -\sqrt{2}\Im \ln \left(\sqrt{2-\sqrt{2}}e^{i\tan^{-1}\frac{-1}{\sqrt{2}-1}} \right) - \frac{(1+\sqrt{2})\pi}{8} \\ & = -\sqrt{2}\Im \ln \left(\sqrt{2-\sqrt{2}}e^{-i\frac{3\pi}{8}} \right) - \frac{(1+\sqrt{2})\pi}{8} \\ & = \frac{3\sqrt{2}\pi}{8} - \frac{(1+\sqrt{2})\pi}{8} \\ & = \frac{(2\sqrt{2}-1)\pi}{8} \end{aligned}

⇒ A × B × C × D × E = 2 × 2 × 1 × 1 × 8 = 32 \Rightarrow A \times B \times C \times D \times E = 2 \times 2 \times 1 \times 1 \times 8 = \boxed{32}

10 Helpful 1 Interesting 0 Brilliant 0 Confused

@Rajorshi Chaudhuri Can you add a link to the inspiration? Thanks!

Calvin Lin Staff - 5 years, 6 months ago

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